Question

If the radius of a right circular cone is halved and height is doubled, then the volume will remain unchanged.

Answer 1

False.

Let the radius of a right circular cone be $r$ and height of a right circular cone be $h$.

Then, volume of the right circular cone $V=\frac{1}{3}\pi r^{2}h$

New radius and heights are,
$r_{new}=\frac{r}{2},h_{new}=2h$

$\begin{array}{cc}\text{New volume of the right circular cone},V_{new}& =\frac{1}{3}\pi {r_{new}}^2{h}_{new}\\ & =\frac{1}{3}\pi {\left(\frac{r}{2}\right)}^2\times \left(2h\right)\\ & =\frac{1}{3}\pi \times \frac{r^2}{4}\times 2h\\ & =\frac{1}{3}\pi \times \frac{r^2}{2}h\\ & =\frac{1}{2}\left(\frac{1}{3}\pi r^{2}h\right)\\ & =\frac{1}{2}V\end{array}$
The new volume is half of the initial volume.